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Vectors question.

  1. Oct 26, 2012 #1
    Hello,

    I'd like to please verify whether my solution to the problem as described below is indeed correct.

    Ship A is traveling east at a constant velocity of 30km/h. It notices ship B 15 km away and 30° from the north. Ship B is traveling at a constant velocity. 20 minutes later ship A notices ship B 45° from the north, and 20 more minutes after that ship B is observed 60° from the north. The distance between ship A and B at that point in time ought to be calculated, including the velocity (magnitude and direction) of ship B.

    Proposed solution:

    Supposing ship A first spots ship B from O, (0,0):

    1) (1/3Va)(OB + 1/3Vb) / |(1/3Va)||(OB + 1/3Vb)| = 1/SQRT(2)

    2) (2/3Va)(OB + 2/3Vb) / |(2/3Va)||(OB + 2/3Vb)| = SQRT(3)/2

    [Va, Vb, OB are vectors; 1/SQRT(2) = cos 45°; SQRT(3)/2 = cos 30°]

    I got Vb = (27.56,15.91), |Vb| = 31.82 km/h, distance between ship A and B 40 minutes later = 12.14 km

    Are these equations correct? Is their solution?
     
  2. jcsd
  3. Oct 26, 2012 #2
    Does 30 degrees from north mean 30 degrees from north towards east or 30 degrees from north towards west?
     
  4. Oct 26, 2012 #3

    haruspex

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    Can you explain the logic behind that equation? It appears to be saying that angle A'OB' is 45 degrees, where A' and B' are the positions after 20 minutes.
     
  5. Oct 27, 2012 #4
    30 degrees from the north means 60 degrees from the east. The movement is only in the east-west plane. That is my assumption from the details in any cass.
     
  6. Oct 27, 2012 #5
    I think there might be two solutions but if we assume 30 degrees from north means towards east I think your numbers are off. Using ruler and compass I solved the problem graphically. I was careful and think my number are off less then 10%. See below,
     

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  7. Oct 27, 2012 #6

    haruspex

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    As I indicated before, I don't think your equations are right. Your equation (1) appears to be saying that angle A'OB' is 45 degrees, where A' and B' are the positions after 20 minutes. It's the angle between OA' and A'B' that's 45 (or 135) degrees.
     
  8. Oct 28, 2012 #7
    Hi,
    For the problem as described above, are the following two equations correct?

    Supposing A = (Va,0)t and B = (15cos60,15sin60) + (Vbx, Vby)t

    Thus:

    20 minutes later:
    A = (1/3Va,0) and B = (15cos60,15sin60)+(1/3Vbx,1/3Vby)

    40 minutes later:
    A = (2/3Va,0) and B = (15cos60,15sin60)+(2/3Vbx,2/3Vby)

    Hence:

    (1) 1/SQRT(2) = (1/3Va,0)*(15cos60+1/3Vbx-1/3Va,15sin60+1/3Vby) / |(1/3Va,0)||(15cos60+1/3Vbx-1/3Va,15sin60+1/3Vby)|

    (2) SQRT(3)/2 = (2/3Va,0)*(15cos60+2/3Vbx-2/3Va,15sin60+2/3Vby) / |(2/3Va,0)||(15cos60+2/3Vbx-2/3Va,15sin60+2/3Vby)|

    Is that correct?
     
  9. Oct 28, 2012 #8
    In case these last equations are also wrong, please suggest any other method/set of equations by which this problem may be approached and solved.
     
  10. Oct 28, 2012 #9

    haruspex

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    Let's approach this geometrically. Let A, B be the initial positions, A', B' the positions after 20 minutes, and A", B" after 40. Spinnor's diagram is useful.
    What angle are we told is 45 degrees?
    What lines border that angle?
    What vectors (treating A as the origin) correspond to the orientations of those lines?
    Your equation (1) appears to be using OA', OB' as those vectors.
     
  11. Oct 28, 2012 #10
    I have made some corrections to my initial equations. They are valid now, wouldn't you agree?
     
  12. Oct 28, 2012 #11
    (1) 1/SQRT(2) = (1/3Va,0)*(15cos60+1/3Vbx-1/3Va,15sin60+1/3Vby) / |(1/3Va,0)||(15cos60+1/3Vbx-1/3Va,15sin60+1/3Vby)|

    simplify -->

    (1) 1/SQRT(2) = (15cos60+1/3Vbx-1/3Va) / |(15cos60+1/3Vbx-1/3Va,15sin60+1/3Vby)|

    Do the same for the other eq.
     
    Last edited: Oct 28, 2012
  13. Oct 28, 2012 #12
    The equations indeed seem correct, but I have not been able to extricate Vbx and Vby. May you please try it yourself and suggest how to go about it?
     
  14. Oct 28, 2012 #13

    haruspex

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    I would start by making the equations simpler to write out. Put x for Vbx etc., multiply out to get rid of the thirds, maybe turn cos 60 and sin 60 into numbers.
    You have two quadratic equations with two unknowns. In general, that could give you a quartic, but maybe you get lucky. Post what you do get.
     
  15. Oct 29, 2012 #14
    Hi,
    It is always possible that I have made some mistakes along the way, but I have got:

    (1) x^2-37.5x -67.5SQRT(3)y-3y^2-787.5=0
    (2) x^2-15x+45SQRT(3)y-y^2-1462.5=0

    Which doesn't get me anywhere. Any advice?
     
  16. Oct 29, 2012 #15
    I did make a mistake, and will repost the correct equations shortly
     
  17. Oct 29, 2012 #16
    Hi,

    I still get:

    (1) x^2 -37.5x -67.5SQRT(3)y -3y^2 -787.5=0
    (2) x^2 -15x -45SQRT(3)y -y^2 -1462.5=0

    Which doesn't get me anywhere. Any advice?
     
  18. Oct 29, 2012 #17

    haruspex

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    If you take the difference of the two equations that will remove the x^2 term, leaving an expression for x in terms of y and y^2. You can use that to substitute for x in one equation. That produces a quartic in y (as I feared). You can look up how to solve quartics. I'm surprised the problem is this hard, but the logic all seems correct.
     
  19. Oct 29, 2012 #18
    There has got to be an easier way. Good for you for not giving up. I think it might make sense to go to a different reference frame. Say we go into the frame of reference of ship A. Will work on this.
     
  20. Oct 29, 2012 #19
    Does this look like a promising route?
     

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  21. Oct 30, 2012 #20

    haruspex

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    Let's go back to the start and look for an easier way.
    Suppose B travels x km E and y km N (could be negative) each 20 minutes.
    What are its co-ordinates (B') after 20 minutes relative to A (A's initial position)?
    What are they relative to A' (A's position after 20 minutes)?
    What can we say about the ratio of the co-ordinates of B' relative to A'?
     
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